Below represents a differential measurement system for a data-acquisition:

enter image description here

As far as I understood, in a differential measurement system neither the outputs of the signals nor the inputs to the differential amplifier is tied to ground. That's why many places I read that this makes it immune to ground noise.

But there is still a measurement ground (AI GND and G) for the system.

My question is about the differential/instrumentation amplifier part:

Does such an amplifier above measures the:

1-) voltage difference between the point A and B directly(without any reference) ?


(voltage difference between A and G) - (voltage difference between B and G) ?

I can already see both above mathematically equivalent. But does sometimes this measurement ground G is also tied to earth ground which might be bouncing(?).

2-) Is the practice of connecting AI GND and G to prevent exceeding common mode voltages? What happens if AI GND and G are not connected and both signal and measurement side ground is not connected to earth?

3-) Should the point G never be tied to the "earth ground" in any side? If it is, what would be the consequences? Or is doesn't matter?

  • \$\begingroup\$ Your initial assumption is wrong. Although not tied, they must be referenced by some reasonable low impedance compared to impedance and level of noise for immunity. There must alway be a CM range that does not exceed the DC limits of CMR of the Instrument Amp. (INA) \$\endgroup\$ Sep 21, 2017 at 17:23
  • \$\begingroup\$ Please tell me is G the circuit ground of the amolifier? And G should not be wired to earth ground in this kind of system? \$\endgroup\$
    – GNZ
    Sep 21, 2017 at 17:34
  • \$\begingroup\$ Study what makes a instrumentation amplifier so special. They are so well designed that they seem to measure directly the difference between A and B with high impedance. However, inside the instrumentation amplifier there is a lot going on. \$\endgroup\$
    – Jot
    Sep 21, 2017 at 18:40
  • \$\begingroup\$ @Jot Im trying to understand how the ground is not involved in subtraction process. Innerly basically as a big picture. Is that something related to the input stage of a diff amplifier? \$\endgroup\$
    – GNZ
    Sep 21, 2017 at 18:44
  • \$\begingroup\$ @by ground I mean the grund of the diff. amp circuit not the "earth" \$\endgroup\$
    – GNZ
    Sep 21, 2017 at 18:55

1 Answer 1


Consider the following differatial amplifier circuit.


simulate this circuit – Schematic created using CircuitLab

Note the input to the op-amp is a voltage offset and is not referenced directly to the system ground. As such you are pushing/pulling current to/from the +/- pins. The op-amp uses that difference to operate regardless of ground.

The key then is that both inputs must have their own common connection. Whether that be through AI ground, or completely isolated.

As for grounding. To answer your second point first, a connection to earth ground is beneficial for a number of reasons including static build up protection and EMI/EMC reasons. A single point connection is preferred such that the connection carries no current.

Connection between AI Ground and GND on the other hand is more complicated. If they are close then they should normally be connected. If on the other hand the AI Part is on the other side of the building and or there are large currents flowing though the ground bar, then differences in the ground level can introduce unwanted effects if you connect them together. It therefor gets complicated quickly and significant amount of time and effort needs to be put in to come up with the best grounding solution for each application.

ADDITION: On re-examining your question I noticed you are using an instrumentation amplifier not a differential amplifier. That changes things since Instrumentation Amplifiers have a buffer stage on each input. For this type of circuit you may need some common reference.

enter image description here

  • \$\begingroup\$ "input to the op-amp is a voltage offset and is not referenced directly to the system ground." What do you mean by "ground" here? Ground of the diff. amp circuit or earth ground? \$\endgroup\$
    – GNZ
    Sep 21, 2017 at 18:17
  • \$\begingroup\$ @161776 I mean V3, the 3.14V source, is floating with respect to everything else. It's a stand alone "two terminal black-box" that generates a difference of voltage across it's outputs. \$\endgroup\$
    – Trevor_G
    Sep 21, 2017 at 18:18
  • \$\begingroup\$ But right after you wrote "both inputs must have their own common connection." But in your example it is floating there is no ground. I couldnt make a connection. \$\endgroup\$
    – GNZ
    Sep 21, 2017 at 18:24
  • \$\begingroup\$ @161776 that is a simplified source though, The common connection between the + and - of the battery is inside the battery. If it's a circuit, obviously the positive and negative signals need to be referenced to each other. \$\endgroup\$
    – Trevor_G
    Sep 21, 2017 at 18:26
  • \$\begingroup\$ The input signals can have common connection but that point is not tied to the diff. amplifier ground right? So it is floating? One more thing the output of the diff. amplifier is with respect to its ground but when it measures/subtracts the inputs i doesn't use its ground in the subtraction process? \$\endgroup\$
    – GNZ
    Sep 21, 2017 at 18:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.